Estonian Math Competitions

Assume that there exist $a_i$ and $a_j$ which are not equal. Let $S = a_1 + \dots + a_{2n}$. Since $S - a_i$ and $S - a_j$ are integers, their difference $(S - a_i) - (S - a_j) = a_j - a_i$ is also an integer. Since $a_j - a_i \neq 0$, and they belong to $\left[-\frac{1}{2}, \frac{1}{2}\right]$, their difference can be only $\pm 1$, this happens when $a_i$ and $a_j$ are $\frac{1}{2}$ and $-\frac{1}{2}$ in any order. Let $a_k$ be any of the given numbers. Since $(S - a_i) - (S - a_k) = a_k - a_i$ is an integer, $a_k$ must also be either $\frac{1}{2}$ or $-\frac{1}{2}$. Hence all numbers $a_i$ are either $\frac{1}{2}$ or $-\frac{1}{2}$. It follows that the sum of any two numbers $a_i$ and $a_j$ is an integer. As we have an even number of them, the sum of all the numbers $S$ is also an integer. But then $S - a_i = S \pm \frac{1}{2}$ cannot be an integer, a contradiction. Therefore all numbers $a_i$ must be equal.